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Given $f^3$ and $f^7$ holomorphic functions, I want to show that $f$ is holomorphic.

Can I say that if $f$ is not holomorphic then $f^3$ is not holomorphic which mean $f$ is must be holomorphic?

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Is $f^n = f \cdot f \cdot \dots \cdot f$ or $f^n = f \circ f \circ \dots \circ f$? –  Alexander Thumm Aug 16 '12 at 10:38
    
To your last bit, no: consider ${}^3\sqrt{x}$. It most often doesn't work to ignore half the given information. –  Kevin Carlson Aug 16 '12 at 10:40
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Even more drastic: take any function $f: \mathbb C \to \{ 1, \zeta, \zeta^2\}$, where $\zeta$ is a nontrivial third root of unity. –  Alexander Thumm Aug 16 '12 at 10:44

1 Answer 1

up vote 8 down vote accepted

As David pointed out, $f(z)=\frac{f^7(z)}{f^6(z)}$ is holomorphic at all points where $f^6(z)\neq0$. Using Riemann's theorem on removable singularities it's enough to show that $\lim_{z\to0}f(z)$ exists. Now, assume that $f^6(z_0)=0$. Since $f^6$ is holomorphic at $z_0$, you can write $f^6(z)=(z-z_0)^kg(z)$ where $g$ is holomorphic and $g(z_0)\neq0$. Also, $f^6(z_0)=0$ implies that $f^7(z_0)=0$, so you can write $f^7(z)=(z-z_0)^mh(z)$. Now consider $f^{42}$: it is holomorphic, as a product of holomorphic and hence $(z-z_0)^{7k}g(z)=(z-z_0)^{6m}h(z)$. Since the order of the zero at $z_0$ is constant, we have $7k=6m$, so $m>k$. Hence you can write $f(z)=(z-z_0)^{m-k}\frac{h(z)}{g(z)}$, and $g$ doesn't vanish at $z_0$, so the limit of $f(z)$ at $z_0$ exists (and is equal to $0$).

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This is very nice:+1. However @David Wallace posted a solution before, which Dennis's reasoning now makes complete. Since I'm the one who pointed out to David his gap, I'd like to ask him to undelete his answer and to just add a little pointer to Dennnis's answer in an edit. –  Georges Elencwajg Aug 16 '12 at 12:46

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